Williamson synthesis involves the reaction of an alkoxide ion with an alkyl halide via an $S_{N}2$ mechanism. To minimize steric hindrance,the alkyl halide should be primary. Therefore,ethanol is converted to bromoethane,and $3-$methylpentan$-2-$ol is converted to its sodium alkoxide.
$1.$ Conversion of ethanol to bromoethane:
$C_{2}H_{5}OH + HBr \to C_{2}H_{5}Br + H_{2}O$
$2.$ Conversion of $3-$methylpentan$-2-$ol to sodium $3-$methylpentan$-2-$oxide:
$CH_{3}CH(CH_{3})CH(OH)CH_{2}CH_{3} + Na \to CH_{3}CH(CH_{3})CH(ONa)CH_{2}CH_{3} + \frac{1}{2}H_{2}$
$3.$ Williamson synthesis reaction:
$CH_{3}CH(CH_{3})CH(ONa)CH_{2}CH_{3} + C_{2}H_{5}Br \to CH_{3}CH(CH_{3})CH(OC_{2}H_{5})CH_{2}CH_{3} + NaBr$
(Product: $2-$ethoxy$-3-$methylpentane)